Let $E$ be a commutative monoid with identity $e$, $S\subseteq E$, and let $S'$ be the submonoid of E generated by $S$.
For every invertible element $x$ of E, $x^{*}$ denotes the inverse of $x$.
Let $R(x,y)$ denote the following relation:
$$\text{there exist }a,b\in E\text{ and }p,q,s\in S'\text{ such that }x=(a,p), y=(b,q)\text{ and }aqs=bps;$$
this is an equivalence relation on the product monoid $E\ \times\ S'$, compatible with its law. Denote the quotient monoid $E\times S'/R$ by $E_{S}$.
For $a\in E$ and $p\in S'$, let $a/p$ denote the equivalence class of $(a,p)$ modulo R. Then, mapping
$$\varepsilon: E\longrightarrow E_{S},\ a\mapsto a/e,$$
is a homomorphism.
For every $a\in E$ and $p\in S'$, we have $a/p=(a/e)(e/p)$: i.e., $a/p=\varepsilon(a)\varepsilon(p)^{*}.$
Therefore, $\varepsilon(E) \cup\varepsilon(S)^{*}$ generates $E_{S}$, where $\varepsilon(S)^{*}$ denotes the set of invertible elements of $\varepsilon(S)$.
This construction is taken from Bourbaki's Algebra. Now Bourbaki goes on to say:
Question: Can someone please explain what is going on in this screenshot? (I understand why $\varepsilon$ is injective.) In particular, I don't understand how one can "identify" $E_S$ with the submonoid of $\Phi$ generated by $E\cup S^*$? This whole business of "identification" is unclear to me: I don't understand what I am exactly able to do logically and formally whenever I am to told to "identify" two objects.
Any help will be appreaciated. Thank you.